Geometry in a circle Can somebody explain to me why the line $DB$ is equal to $\sin(\alpha + \beta)$. I haven't been able to figure it out. My image (which I would post if I could) is located here: 
For example,
if I try to use the law of cosines then,
$$(\cos \alpha)^2 + (\cos \beta)^2 - 2(\cos \alpha) (\cos \beta) (\cos(\alpha + \beta)) = DB$$
which does not obviously equal $\sin(\alpha + \beta)$.
 A: The circle is a circle with unit diameter. Recall the sine rule which goes $$\dfrac{a}{\sin(A)} = \dfrac{b}{\sin(B)} = \dfrac{c}{\sin(C)} = 2R$$
In you case, $2R = 1$, and $\angle A = \alpha + \beta$. And you use the fact that for a cyclic quadrilateral the product of the diagonals is the sum of the product of opposite sides, which is called Ptolemy's theorem, i.e.
$$BD \cdot AC = BC \cdot AD + AB \cdot CD$$
to conclude that
$$\sin(\alpha + \beta) \times 1 = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)$$
A: The reason why that line has length $\sin(\alpha+\beta)$ is the same as the reason why the line of length $\sin\alpha$ has that length: In a circle of unit diameter, the length of the chord connecting the endpoints of an arc is the sine of an angle whose vertex is on the circle and whose rays pass through the endpoints of the arc.
A: (I use $a$ and $b$ instead of $\alpha$ and $\beta$ because I am lazy.)
\begin{align}
& (\cos a)^2 + (\cos b)^2 - 2(\cos a) (\cos b) (\cos(a + b)) \\
= {} & (\cos a)^2 + (\cos b)^2 - 2(\cos a)(\cos b)(\cos a \cos b - \sin a \sin b) \\
=  {} & (\cos a)^2 + (\cos b)^2 - 2(\cos a)^2(\cos b)^2
+ 2 (\cos a)(\cos b)(\sin a)(\sin b) \\
= {} & (\cos a)^2(1-(\cos b)^2) + (\cos b)^2(1-(\cos a)^2)
+ 2(\cos a)(\cos b)(\sin a \sin b) \\
= {} & (\cos a)^2 (\sin b)^2 + (\cos b)^2 (\sin a)^2
+ 2(\cos a)(\cos b)(\sin a \sin b) \\
= {} & (\cos a \sin b + \cos b \sin a)^2 \\
= {} & (\sin(a+b))^2.
\end{align}
So I think the expression you have
is for $DB^2$, not $DB$, but, other than that,
your expression was correct - just not simplified.
